1. Assembly Language and Computer Architecture Using C++ and Java
5/27/05

2. Chapter 1
Number Systems

3. Would you trust a medical doctor who did not know the location of the human appendix? Not likely. A doctor must know the structure and operation of the human body.

4. Similarly, a computer expert must know the structure and operation of computers.
Providing a clear, concrete, and thorough view of the structure and operation of computers is our objective.

5. Work at this level
Understand at this level

6. Positional number systems
Decimal: base 10
Binary: base 2
Hexadecimal: base 16

7. Decimal
Ten symbols: 0, 1, 2, …, 9
Weights change by a factor of 10 from one position to the next.
Shifting the decimal point changes the value by a factor of 10 for each position shifted.
Each symbol is called a “digit”

8. 25.4 decimal

9. 25.4 decimal =

10. Binary Two symbols: 0 and 1
Weights change by a factor of 2 from one position to the next.
Shifting the binary point changes the value by a factor of 2 for each position shifted.
Each symbol is called a “bit”.

11. binary

12. Most and least significant bits
MSB lsb

13. Hexadecimal Sixteen symbols: 0, 1, 2, …, 9, A, B, C, D, E, F
Weights change by a factor of 16 from one position to the next.
Shifting the hexadecimal point changes the value by a factor of 16 for each position shifted.
Each symbol is called a “hex digit”
A = 10 decimal B = 11 decimal
C = 12 decimal D = 13 decimal
E = 14 decimal F= 15 decimal

14. 1CB.8 hex

15. Memorize this table

16. Complement of a symbol s in a base n system
Binary: Complement of 1 is 1 – 1 = 0
Decimal: Complement of 3 = 9 – 3 = 6
Hex: Complement of 4 = 15 – 4 = 11 = B

17. Byte Sequence of 8 bits Each bit has two possibilities: 0 or 1
Thus, there are 2x2x2x2x2x2x2x2 = 28 distinct patterns
Exercise: Bytes are also called Octals. Why do we have two names?

18. Example: 128 distinct numbers are representable with 7 bits but the number 128 requires 8 (7+1) bits to be represented

19. Arithmetic with positional numbers
Rules are essentially the same for all bases

20. Arithmetic in decimal

21. Adding in binary

22. Subtraction in binary

23. Addition in hex

24. Subtraction in hex

25. Converting to base n
Integer part: Repeatedly divide by n. Remainders are the digits of the base n number (in reverse order).
Fractional part: Repeatedly multiply by n. Whole parts are the digits of the base n number ( times n = move decimal point to the right one place).

26. Each remainder is a digit

27. Convert 11 decimal to binary
11 decimal = 1011 binary

28. Convert 30 decimal to hex
30 decimal = 1E hex

29. Fractional Conversion:
Each whole part is a digit (strip whole part of product after each multiplication by 10)
.43
x 10
1st digit
2nd digit

30. Convert .6875 decimal to binary
.6875 decimal = binary

31. Repeating fraction
.1 decimal = binary

32. Converting between binary and hex
Very easy to do
The ease of conversion is the reason why hex is often used as a shorthand representation of binary.
To do conversion, you need to know the binary-hex equivalents for the numbers 0 to 15.

33. Binary to hex

34. Hex to binary

35. Horner’s method
Efficient technique for converting to a new number base
(just multiple by that base)

36. Inefficient evaluation of 2CD5

37. Use Horner’s rule to convert 2CD5

38. Horner’s rule written horizontally

39. Horner’s Rule Algorithm
Get keystroke;
While (keystroke is digit) { N = N * 10 + keystroke; }

40. Convert 38 to binary using Horner’s rule
Shift right == *2
corresponds to position
of multiplier

41. Signed binary numbers Sign-Magnitude Two’s Complement One’s Complement
Excess-n

42. Sign-magnitude
Exercise: Add these two numbers???

43. Sign-magnitude representation not good for computers
Adding sign-magnitude requires sign analysis: Like signs, then add magnitudes Unlike signs, then subtract magnitudes

44. Two’s complement
The two’s complement of the number M is the number which yields 0 when added to M.
Just the definition of negative

45. Note this behavior So 1111…1 is just 1 less than 0. Remember that.
what number is this?
So 1111…1 is just 1 less than 0. Remember that.

46. Simply “flipping” the bits does not yield the two’s complement.
But you are not far from 0; just add 1

47. Getting the two’s complement
Simply flipping the bits does not yield the two’s complement—it yields all 1’s.
But adding 1 to all 1’s yields all 0’s.
So flipping the bits and adding 1 should yield the two’s complement.

48. Flip bits and add 1

49. Two’s complement of 1

50. 3-bit two’s complement numbers

51. Terminology

52. Positive result is in true form. Negative result in complement form.

53. One’s complement Like two’s complement, but don’t add 1
Requires addition of end-around carry
= +1
= -1
skip this in class

54. Adding one’s complement numbers
Processing end-around carry is hard for computers

55. One’s complement has two representations of 0: and This feature is not good for computers.

56. Excess-n representation
n is added to a number to get its excess-n representation.
n is subtracted from an excess-n number to get its value.

58. When adding two excess-n numbers, must subtract n

59. Subtracting by two’s complement addition: subtracting +3 from 5

60. Use two’s complement to subtract unsigned numbers

61. Range of unsigned numbers
n bits means 2n patterns
Thus, range is 0 to 2n – 1
n = 8, range is 0 to 28 – 1 ( 0 to 255)

62. Range of two’s complement numbers
Positive numbers have msb = 0. Thus, there are n – 1 bits that can be set.
Positive range is 0 to 2n-1 - 1
Negative numbers have msb = 1. Thus, there are n – 1 bits that can be set
Negative range is -2n-1 to -1.
Total range: -2n-1 to 2n-1 – 1.

63. 2’s Complement Problem:
Since there are the same number of +ve and –ve numbers, why is the largest –ve number (-2n-1) while the largest +ve number appears to be 1 less: (2n-1 -1) -1
2n-1 - 1
2n-1
same distance

64. memorize
this

65. Danger in using char to represent signed number
When extending a variable of type char to a longer format, extension may be signed number extension OR unsigned number extension.
Type of extension depends on the compiler.

67. Signed overflow
Overflow never occurs when adding numbers of unlike signs. -1 b
2n-1 - 1
2n-1 a
overflow
region
overflow
region
a + b

68. Signed overflow
Overflow is indicated if the carry into the msb column differs from the carry out of the msb column. b a -1 b a
2n-1 - 1
2n-1
overflow
region
overflow
region
also possible
a + b

69. MSBs
(0)
(1)

71. Flags V is the signed overflow flag S is the sign flag
C is the carry flag

72. Unsigned overflow on an addition indicated by a carry out

73. Another type of unsigned overflow
Another type of unsigned overflow. Indicated by a borrow into msb on a subtraction. 1

74. Unsigned overflow during a subtraction
Indicated by a borrow into the msb.
But computers do not subtract using the borrow technique.
Computers subtract using the two’s complement technique.
Carry out occurs if and only if a borrow in would not have occurred had the borrow technique been used.

75. Unsigned overflow test
“borrow in” with the borrow technique iff no “carry out” with two’s complement technique.
Addition:
set the carry flag to the carry out.
Subtraction:
set the carry flag to the complement of the carry out.
So carry flag functions as carry/borrow flag.
Addition or subtraction: a 1 in the carry flag indicates unsigned overflow.

76. borrow 2s-comp (1) 0001 0001 - 0010 +1110 1111 (0) 1111 (0) 0010 0010
Examples
borrow s-comp
(1)
(0) 1111
(0)
(1) 1111

77. Flip carry out on a subtraction
0: on addition
1: on subtraction

78. Analyzing two’s complement numbers
2n is an (n+1)-bit number. For example,
24 = (5 bits)
2n – 1 is an n-bit number containing all 1’s.
For example, 24 – 1 = 1111, but to make it positive we put it in a 5-bit number,

79. Analyzing two’s complement numbers
To flip an n-bit number, first subtract from all 1’s:
1111
- 0101
(0101 flipped)
All 1’s is 2n -1. Thus, the n-bit two’s complement of x is 2n -1 – x + 1 = 2n - x
(N – 1) - number

80. Analyzing two’s complement numbers
2n – x (2’s comp of x) x
2n + 0
Carry out of leftmost column

81. Analyzing two’s complement numbers
In mathematics, adding two positives yields a positive; adding two negatives yields a negative; adding a positive and a negative yields either a positive or negative depending on which of the two numbers added has the larger magnitude.
Two’s complement numbers behave in the same way.

82. Analyzing two’s complement numbers
Adding two positive two’s complement numbers: a b
a + b
The n-bit result equals the true value of a + b
as long as overflow does not occur. If it occurs it turns the msb on making the number negative

83. Analyzing two’s complement numbers
Adding two negative two’s complement numbers:
2n – c (two’s complement of c)
2n – d (two’s complement of d)
2n + 2n - (c + d) (two’s complement of c+d)
Lost
Notice that the sum of two negatives is negative.

84. Analyzing two’s complement numbers
Adding a positive and a negative two’s complement number, a < d: a
2n – d (two’s complement of d)
2n - (d - a) (two’s complement of d - a)
Notice the result is negative.

85. Analyzing two’s complement numbers
Adding a positive and a negative two’s complement number, a >= c a
2n – c (two’s complement of c)
2n + a - c
2n + a - c > 2n so needs n+1 bits
Lost
Notice the result is positive.

86. Full Adder
1  carry in 1 +0
------
0  1 carry out

88. Gates
Simple circuits usually with two input lines and one output line, each carrying one bit. The output at any time depends on the inputs at that time.

89. negates the
original meaning

90. XOR gate
Complementing gate

91. Complementing action of the XOR gate

93. Zero detecting gate Outputs 1 if all inputs are 0
NOR gate
Zero detecting gate
Outputs 1 if all inputs are 0

95. Subtracting a number from itself
Using the borrow technique, a borrow would never occur.
Thus, using the two’s complement technique, a carry out should ALWAYS occur.

96. A carry out should always occur when subtracting a number from itself.

97. Subtracting 0 from 0 is not an exception

98. Sign of true result
Needed to determine the result of a signed comparison.
The S flag has the sign of the true result of a two’s complement addition only if overflow does NOT occur.
To get true sign, use S XOR V.

99. Getting the sign of the true result

100. Scientific notation Significand x power of 10
where significand is the number in which the decimal point is to the immediate right of the leftmost non- zero digit.
done in the lab

101. 154 in scientific notation is 1.54 x 102

102. Binary scientific notation
= x 23

103. Floating point format for +1.11011 x 23
Its sign is +.
is called the significand
is called the fractional part
3 is the exponent
In floating-point format, store the sign (0 for +, 1 for -), the exponent in excess-127, and the fractional part.

104. Floating-point format for 1
Floating-point format for x 23 Note that the bit to the left of the binary point does not appear in the floating point format. It is called the hidden bit.

105. Normalization
Shifting a number so that the binary point is in the correct place for floating-point format (to the immediate right of the leftmost 1).

106. Special values

107. Denormal number Exponent = -126 and hidden bit = 0

108. Computational error Floating-point computations are not exact.
Errors can grow to the point where the final results of a computation may be meaningless.

109. The sum computed by the first program on the next slide is not 1.

111. The first program on the next slide is an infinite loop.

113. Round-off error

115. The program on the next slide displays “equal”.

117. Associative operator Order of performing operations does not matter.
(a + b) + c = a + ( b + c)

118. The programs on the next slide output different sums.

120. Suppose x and y are true values, and X and Y, respectively, are their computed values that include small errors.

121. X = x + 0. 0001 Y = y -0. 0001 Now compute X – Y: X - Y = x + 0
X = x Y = y Now compute X – Y: X - Y = x – (y – ) = (x – y) The absolute error (0.0002) will be large relative to the true result (x - y) if x and y are nearly equal.